Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

When 100's digit is 8 ...the no. of digits is 8X5 =40 No . of cases when units digit is 1= 8 Now, there are 5 odd digits that can appear in units digits . therefore 8 X 5 =40

100's digit is 9 ...the no. of digits is 8 X 4 = 32 In this case the odd no. 9 is in 100'2 digit. Therefore it is not available for units digit. Hence the no. of add distinct digits is 8X4 =32

100s digit can be 8 or 9....hence total 2. 10's digit can be any number from 0 to 9 but excluding the digit that has been selected for 100s digit....hence a total of 9 ways.

unit digit can be 1,3,5,7 or 9....but 9 would be already selected previously, hence a total of 4.

Its 72. Odd nubmer in the last digit- 1, 3, 5, 7, 9 for the 800 series. Since 3# are different, 0, 2, 4, 6, 1,3,5,7,9 can be the middle digit. For the 800series, there will be 5x4+5x4=40 numbers

Odd number in the last digit- 1, 3, 5, 7 for the 900 series. Since 3# are different, 2, 4, 6, 8,1,3,5,7 can be the middle digit. For the 900series, there will be 4x5=4x3=32 numbers.

How many odd three-digit integers greater than 800 are there such that all their digits are different?

* 40 * 56 * 72 * 81 * 104

NUMBERS are 8XY and 9XY:

a. for 8XY:

X can odd or even integer other than 8 for integers between 800 and 900:

if X is even, y can be all 5 odd integers. so no of ways = 4*5 = 20 if X is odd, y can be all remaining 4 odd integers. so no of ways = 5*4 = 20

b. for 9XY:

X can odd or even integer other than 9 for integers between 900 and 1000:

if X is even, y can be all 4 odd integers except 9. so no of ways = 5*4 = 20 if X is odd, y can be all remaining 3 odd integers. so no of ways = 4*3 = 12

is something wrong with this question? I think the total number of such digits is 80. It cannot be 72. Aren't all the solutions posted above counting numbers with similar digits? The questions states how many odd numbers with different digits.

There will be 45 numbers starting with 8 and 35 starting with 9. This gives a total of 80?

is something wrong with this question? I think the total number of such digits is 80. It cannot be 72. Aren't all the solutions posted above counting numbers with similar digits? The questions states how many odd numbers with different digits.

There will be 45 numbers starting with 8 and 35 starting with 9. This gives a total of 80?

How many odd three-digit integers greater than 800 are there such that all their digits are different?

In the range 800 - 900: 1 choice for the first digit: 8; 5 choices for the third digit: 1, 3, 5, 7, 9; 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

1*5*8 = 40.

In the range 900 - 999: 1 choice for the first digit: 9; 4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit); 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

Let the 1st digit be 8. The second digit can be either odd (5 options) or even (4 options). The third digit has 4 options (if the 2nd is odd) and 4 options (if the 2nd is even). So you get: 1x5x4 + 1x4x5 = 40

Let the 1st digit be 9. The second can be either odd (4 options) or even (5 options). The third digit has 3 option (if the 2nd is odd) and 4 options (if the 2nd is even). So you get: 1x4x3 + 1x5x4 = 32

i'm still not getting the logic..i find myself convincing myself of different things for examplle

for 800s first digit can be 8 only 2nd digit can be 0,1,2,3,4,5,6,7,9 3rd digit can be 1,3,5,7,9 1x9x5

for 900s 1st 9 2nd 0,1,2,3,4,5,6,7,8 3rd 1,3,5,7 1x9x4

how is the answer not 45 + 36 ?

This approach is not correct. If you say that (in the range 800-900) for the second digit you have 9 choices, then for the third digit you'll have sometimes 5 choices (in case you choose even for the second) and sometimes 4 choices (in case you choose odd for the second), so you can not write 1*9*5.

Correct way would be to count the choices for the third digit first - 5 choices (1, 3, 5, 7, 9) and only then to count the choices for the second digit - 8 choices (10-first digit-third digit=8) --> 1*5*8=40.

Similarly for the range 900-999: 1 choice for the first digit: 9; 4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit); 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.
_________________

Re: How many odd three-digit integers greater than 800 are there [#permalink]

Show Tags

16 May 2015, 20:58

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: How many odd three-digit integers greater than 800 are there [#permalink]

Show Tags

13 Nov 2016, 02:29

study wrote:

How many odd three-digit integers greater than 800 are there such that all their digits are different?

A. 40 B. 56 C. 72 D. 81 E. 104

Answer is D, between 801 to 999 total 199 numbers are there. a. Total odd numbers are(199/2+1) 100, because first and last numbers are odd, b. we have number from like (881, 883, 885, 887 & 889) - 5 numbers repeating 8 two times c. Like 9 repeating another 5 Numbers(991, 993, 995, 997 & 999) d. We have number like 811, 833, 855, 877 & 899, 5 Numbers e. We have another numbers like 911, 933, 955, 977 - numbers(999 already removed) Total numbers is = 100-5-5-5-4 = 81 numbers

Please correct me if I am wrong. Please remove the question if i am correct. Save precious time of aspirants by removing useless questions

How many odd three-digit integers greater than 800 are there such that all their digits are different?

A. 40 B. 56 C. 72 D. 81 E. 104

Answer is D, between 801 to 999 total 199 numbers are there. a. Total odd numbers are(199/2+1) 100, because first and last numbers are odd, b. we have number from like (881, 883, 885, 887 & 889) - 5 numbers repeating 8 two times c. Like 9 repeating another 5 Numbers(991, 993, 995, 997 & 999) d. We have number like 811, 833, 855, 877 & 899, 5 Numbers e. We have another numbers like 911, 933, 955, 977 - numbers(999 already removed) Total numbers is = 100-5-5-5-4 = 81 numbers

Please correct me if I am wrong. Please remove the question if i am correct. Save precious time of aspirants by removing useless questions

Military MBA Acceptance Rate Analysis Transitioning from the military to MBA is a fairly popular path to follow. A little over 4% of MBA applications come from military veterans...

Best Schools for Young MBA Applicants Deciding when to start applying to business school can be a challenge. Salary increases dramatically after an MBA, but schools tend to prefer...

Marty Cagan is founding partner of the Silicon Valley Product Group, a consulting firm that helps companies with their product strategy. Prior to that he held product roles at...